Question

12．若方程組$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解為$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$，那么方程組$\left\{\begin{array}{l}{2{a}_{1}x+3{b}_{1}y=5{c}_{1}}\\{2{a}_{2}x+3{b}_{2}y=5{c}_{2}}\end{array}\right.$的解為$\left\{\begin{array}{l}{x=5}\\{y=5}\end{array}\right.$．

Answer 1

分析 根據方程組$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解為$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$，代入變形后即可建立與方程組$\left\{\begin{array}{l}{2{a}_{1}x+3{b}_{1}y=5{c}_{1}}\\{2{a}_{2}x+3{b}_{2}y=5{c}_{2}}\end{array}\right.$的關系，從而可以解答本題．

解答 解：∵方程組$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解為$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$，
∴$\left\{\begin{array}{l}{2{a}_{1}+3{b}_{1}={c}_{1}}\\{2{a}_{2}+3{b}_{2}={c}_{2}}\end{array}\right.$
∴$\left\{\begin{array}{l}{2{a}_{1}×5+3{b}_{1}×5={c}_{1}×5}\\{2{a}_{2}×5+3{b}_{2}×5={c}_{2×5}}\end{array}\right.$
即$\left\{\begin{array}{l}{2{a}_{1}×5+3{b}_{1}×5=5{c}_{1}}\\{2{a}_{2}×5+3{b}_{2}×5=5{c}_{2}}\end{array}\right.$，
∴方程組$\left\{\begin{array}{l}{2{a}_{1}x+3{b}_{1}y=5{c}_{1}}\\{2{a}_{2}x+3{b}_{2}y=5{c}_{2}}\end{array}\right.$的解為：$\left\{\begin{array}{l}{x=5}\\{y=5}\end{array}\right.$，
故答案為：$\left\{\begin{array}{l}{x=5}\\{y=5}\end{array}\right.$．

點評 本題考查二元一次方程的解，解題的關鍵是利用數學中轉化的思想建立已知條件與所求問題之間的關系．